Question

When a plastic thin film of refractive index $1.45$ is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is $5890\,\mathring{A}$, will be

• A. $6.544 \times 10^{-4} cm $
• B. $6.544 \times 10^{-4} m $
• C. $6.54 \times 10^{-4} cm $
• D. $6.5 \times 10^{-4} cm $

Answer 1

Answer: (A)

$X _{0}=\frac{\beta}{\lambda}(\mu-1) t $
$\Rightarrow 5 \beta=\frac{\beta(0.45) t }{5890 \times 10^{-10}}$
$t =\frac{5 \times 5890 \times 10^{-10}}{0.45}$
$=6.544 \times 10^{-4} \,cm$